# Construction of Fourth Apotome

## Theorem

In the words of Euclid:

To find the fourth apotome.

## Proof Let $A$ be a rational straight line.

Let $BG$ be commensurable in length with $A$.

Therefore, by definition, $BG$ is also rational.

Let $DF$ and $FE$ be square numbers such that $DE$ does not have to either $DF$ or $FE$ the ratio that a square number has to another square number.

Using Porism to Proposition $6$ of Book $\text{X}$: Magnitudes with Rational Ratio are Commensurable, let it be contrived that:

$DE : EF = BG^2 : GC^2$
$BG^2$ is commensurable with $GC^2$.

But $BG^2$ is rational.

Therefore $GC^2$ is rational.

Therefore $GC$ is rational.

We have that $DE : EF$ is not the ratio that a square number has to another square number.

Therefore $BG^2 : GC^2$ is not the ratio that a square number has to another square number.

$BG$ is incommensurable in length with $GC$.

Both $BG$ and $GC$ are rational.

Therefore $BG$ and $GC$ rational straight lines which are commensurable in square only.

Therefore $BC$ is an apotome.

It remains to be shown that $BC$ is a fourth apotome.

Let $H$ be a straight line such that $H^2 = BG^2 - GC^2$.

We have that:

$DE : EF = BG^2 : GC^2$
$ED : EF = GB^2 : H^2$

So $ED$ does not have to $EF$ the ratio that a square number has to another square number.

Therefore neither does $GB^2$ have to $H^2$ the ratio that a square number has to another square number.

$BG$ is incommensurable in length with $H$.

We have that:

$BG^2 = GC^2 + H^2$

Therefore $BG^2$ is greater than $GC^2$ by the square on a straight line which is incommensurable in length with $BG$.

Also, the whole $BG$ is commensurable in length with the rational straight line $A$.

Therefore, by definition, $BC$ is a fourth apotome.

$\blacksquare$

## Historical Note

This proof is Proposition $88$ of Book $\text{X}$ of Euclid's The Elements.